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#### Also sprach Zarathustra

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- Jan 31, 2012

- 54

$$f(z)=\left\{\begin{matrix}e^{-\frac{1}{z^4}}\ \ \ \text{if z not 0 }\\ 0 \ \ \ \ \ \ \ \text{if z not 0 }\end{matrix}\right.$$

Thank you.

- Thread starter Also sprach Zarathustra
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- Thread starter
- #1

- Jan 31, 2012

- 54

$$f(z)=\left\{\begin{matrix}e^{-\frac{1}{z^4}}\ \ \ \text{if z not 0 }\\ 0 \ \ \ \ \ \ \ \text{if z not 0 }\end{matrix}\right.$$

Thank you.

That should be 0 if z is 0 correct?

$$f(z)=\left\{\begin{matrix}e^{-\frac{1}{z^4}}\ \ \ \text{if z not 0 }\\ 0 \ \ \ \ \ \ \ \text{if z not 0 }\end{matrix}\right.$$

Thank you.

$\displaystyle u = \exp\left(\frac{x^4+y^4-6x^2y^2}{x^8+y^8+4x^6y^2+6x^4y^4+4x^2y^6}\right) \cos \left(\frac{4xy^3-4x^3y}{x^8+y^8+4x^6y^2+6x^4y^4+4x^2y^6}\right)$

$\displaystyle v = \exp\left(\frac{x^4+y^4-6x^2y^2}{x^8+y^8+4x^6y^2+6x^4y^4+4x^2y^6}\right) \sin \left(\frac{4xy^3-4x^3y}{x^8+y^8+4x^6y^2+6x^4y^4+4x^2y^6}\right)$

I then used Mathematica to check the derivatives and they are satisfied.

Then you would only have to check the limit.

If you want, I can send you the mathematica nb file.

CORRECTION

I just noticed I forgot the negative sign. Everything should be multiplied through by negative 1.

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- Feb 7, 2012

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I think that the point of this question is to see what happens at the origin (see here).Explain me please how to show that C.R conditions are holds.

$$f(z)=\begin{cases}e^{-1/z^4}&(z \ne 0)\\ 0 & (z = 0) \end{cases}$$

For every nonzero value of $z$, the function $f(z)$ is analytic in a neighbourhood of $z$ and therefore satisfies the C–R equations. The surprising thing is that $f(z)$ also satisfies the C–R equations at $z=0$, despite not being analytic (not even continuous) there.

To see that, write $z=x+iy$ and $f(z) = f(x,y) = u(x,y) + iv(x,y)$. Then $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{(\text{re }f(x,0)) - (\text{re }f(0,0))}{x} = \lim_{x\to0}\frac{\text{re }e^{-1/x^4}-0}x.$$

That limit is 0. Similar calculations show that $\frac{\partial v}{\partial x}$, $\frac{\partial u}{\partial y}$ and $\frac{\partial v}{\partial y}$ are all 0 at the origin, and therefore the C–R equations are satisfied there.

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- Jan 31, 2012

- 54

I think that the point of this question is to see what happens at the origin (see here).

For every nonzero value of $z$, the function $f(z)$ is analytic in a neighbourhood of $z$ and therefore satisfies the C–R equations. The surprising thing is that $f(z)$ also satisfies the C–R equations at $z=0$, despite not being analytic (not even continuous) there.

To see that, write $z=x+iy$ and $f(z) = f(x,y) = u(x,y) + iv(x,y)$. Then $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{(\text{re }f(x,0)) - (\text{re }f(0,0))}{x} = \lim_{x\to0}\frac{\text{re }e^{-1/x^4}-0}x.$$

That limit is 0. Similar calculations show that $\frac{\partial v}{\partial x}$, $\frac{\partial u}{\partial y}$ and $\frac{\partial v}{\partial y}$ are all 0 at the origin, and therefore the C–R equations are satisfied there.

You are right, about the purpose of this question!

I still have troubles make $f(z)$ into the form $f(x,y)=u(x,y)+v(x,y)$.

$$f(x,y)=e^{x^4-6x^2y^2-y^4}e^{4i(xy^3-x^3y)}$$

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- Feb 7, 2012

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In fact, you do not need to find explicit formulas for $u(x,y)$ and $v(x,y)$. To evaluate $\displaystyle\frac{\partial u}{\partial x}$ at the origin, you have to go back to the definition of the derivative $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{(\text{re }f(x,0)) - (\text{re }f(0,0))}{x}$$ (as in my previous comment). That limit is equal to $$\lim_{x\to0}\frac{\text{re }e^{-1/x^4}-0}x.$$ But $e^{-1/x^4}$ is already real, so its real part is itself. Therefore $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{e^{-1/x^4}}x.$$ That is a purely real limit, which is equal to 0 (essentially because exponential convergence is more powerful than polynomial convergence).I still have troubles make $f(z)$ into the form $f(x,y)=u(x,y)+v(x,y)$.

$$f(x,y)=e^{x^4-6x^2y^2-y^4}e^{4i(xy^3-x^3y)}$$

The key idea here is that in order to evaluate the partial derivatives of $f(z)$ at the origin, you only need to evaluate $f(z)$ at points on the coordinate axes. But $f(z)$ is real at all such points. Hence $u$ is equal to $f$, and $v$ is equal to 0, at all such points.